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Local uniqueness of an x₀-minimum-norm solution under TCC

Statement

Let X,Y be real Hilbert spaces, let U⊆X be open, and let F:U→Y be Fréchet differentiable. Let x0∈X, y∈Y, and let x†∈U be an x0-minimum-norm solution of F(x)=y on U. Suppose that for some δ>0 the closed ball B‾δ(x†) lies in U, the local tangential cone condition holds at x† on radius δ with a constant 0≤η<1, and

x†−x0∈ker⁡(F′(x†))⊥.

Then every x∈B‾δ(x†) with F(x)=y and x≠x† satisfies

∥x−x0∥X>∥x†−x0∥X.

Consequently, x† is the unique x0-minimum-norm solution among the exact solutions in B‾δ(x†).

Facts & Assumptions

Given: Real Hilbert spaces X,Y, an open set U⊆X, a Fréchet differentiable map F:U→Y, x0∈X, y∈Y, and an x0-minimum-norm solution x†∈U of F(x)=y. For some δ>0, B‾δ(x†)⊆U and the local tangential cone condition holds there with 0≤η<1. Also assume x†−x0∈ker⁡(F′(x†))⊥.

[F1]

Under the local tangential cone condition, (1−η)∥F(x+h)−F(x)∥Y≤∥F′(x)h∥Y≤(1+η)∥F(x+h)−F(x)∥Y for all admissible h (Tangential cone control compares derivative and data residuals).

[F2]

For a linear map T:V→W, its kernel is ker⁡T:={v∈V:T(v)=0W} (Kernel and image of a linear map).

[F3]

For a subset S of an inner-product space, S⊥:={v:⟨v,s⟩=0 for every s∈S} (Orthogonality and the orthogonal complement).

[F4]

If u,v are orthogonal in an inner-product space, then ∥u+v∥2=∥u∥2+∥v∥2 (Pythagoras and finite orthogonal sums).

[F5]

The induced Hilbert norm is nonnegative and vanishes only at the zero vector (The induced length is a norm).

[A1]

x† is an exact solution and globally minimizes distance to x0 among solutions on U (x₀-minimum-norm solutions of nonlinear equations).

[A2]

The TCC holds at x† on the stated closed ball, with 0≤η<1 (Local tangential cone condition).

[A3]

x†−x0 is orthogonal to every vector in ker⁡(F′(x†)) (Orthogonality and the orthogonal complement).

Proof

technique · direct
1.1givenA2F1F2choosealgebra

Choose any x∈B‾δ(x†) with F(x)=y and x≠x†, and put h:=x−x†. Applying [F1] at x† gives ∥F′(x†)h∥Y≤(1+η)∥F(x)−F(x†)∥Y=0. Hence F′(x†)h=0, so [F2] gives h∈ker⁡(F′(x†)); also h≠0.

2.1step 1.1A1A3F3F4F5algebra∎

By [A3] and [F3], x†−x0 is orthogonal to h. Applying [F4] to x†−x0 and h yields ∥x−x0∥X2=∥x†−x0∥X2+∥h∥X2>∥x†−x0∥X2, since [F5] and h≠0 imply ∥h∥X>0. The norms are nonnegative, so the strict inequality of squares gives the strict norm inequality. With [A1], this proves uniqueness of the x0-minimum-norm solution among the local exact solutions.

Remarks

The conclusion is local minimum-norm uniqueness, not injectivity of F: other exact solutions may lie in the ball, but each has strictly larger distance to x0.

Depends on

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